If I have a flat family $f \colon X \to T$ such that some fiber is (locally) a complete intersection, does that imply that there is an open set $U$ in $T$ such that the fibers above $U$ are (locally) complete intersections?

In general, what kind of intuition should one have about which properties are "open" with respect to flat families?

2 Answers
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EGA IV$_4$, 19.3.8 (and 19.3.6); this addresses openness upstairs without properness, and (as an immediate consequence) the openness downstairs if $f$ is proper (which I assume you meant to require).

The general intuition is that openness holds upstairs for many properties, and so then holds downstairs when map is proper. As for proving openness upstairs, the rough idea is to first prove constructibility results, and then refine to openness by using behavior under generization. But it's a long story, since there are many kinds of properties one can imagine wanting to deal with. These sorts of things are developed in an extraordinarily systematic and comprehensive manner in EGA IV$_3$, sections 9, 11, 12 (especially section 12 for the niftiest stuff).

I just wanted to point out that the situation for globally complete intersections is very different; the property of being a globally complete intersection is not open. For example, consider $(\mathbb{P}^2)^9$. This is an $18$-dimensional irreducible space; the locus of $9$-tuples of points which can be written as a complete intersection of two cubics only has dimension $16$, and thus is not open.

Here is a nice trick question: Take a family of $9$-tuples of points in $\mathbb{P}^2$, indexed by $t \in \mathbb{A}^1$, which are the intersection of two cubics for $t=0$ but not for generic $t$. Cone this family to get a family of $1$-dimensional subschemes of $\mathbb{P}^3$. Why is this not an example of a singularity which is a locally complete intersection deforming to a singularity which is not?